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olga55 [171]
2 years ago
12

Find the circumference of the

Mathematics
1 answer:
WITCHER [35]2 years ago
6 0
The answer is abc deft hi
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Find the x will name brainlist
Alenkasestr [34]

\huge \boxed{\mathbb{QUESTION} \downarrow}

  • Find x.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

To solve these type of problems you need to use the pythagoras theorem ⇨ Hypotenuse² = Base² + Altitude².

Here,

  • Altitude = 1.6 cm.
  • Base = 1.2 cm
  • Hypotenuse = x

Now, let's solve for x.

Hypotenuse² = Base² + Altitude²

x² = (1.2)² + (1.6)²

x² = 1.44 + 2.56

x² = 4

x = √4

x = <em><u>2</u></em><em><u>.</u></em>

  • So, the value of x is <em><u>2</u><u> </u><u>cm.</u></em>

<h3><u>NOTE</u><u> </u><u>:</u><u>-</u></h3>
  • Pythagoras theorem can be used only in the cases of right-angled triangles. Here, it's given that the triangle is right angled so we can use this theorem.
  • To solve the squares if decimals, take them as whole numbers & then just add the decimal points. For example, ⇨ for (1.2)², take it as 12² , then multiply 12 by 12, you'll get 144. Now, add the decimal place accordingly ⇨ 1.44 . So, (1.2)² = 1.44.
8 0
3 years ago
Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

8 0
3 years ago
HELP PLZ PLZ ITS DUE IN 30 MINUTES
Bad White [126]

Answer:

wouldn't that be...24x²?

3 0
2 years ago
Help me pleasee :) :)
andreyandreev [35.5K]
I know that one is obtuse and acute
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Bill made 5 gallons of fruit punch. If 1/4 of the punch was cranberry juice, how much cranberry juice did he use?
Temka [501]
1.25
---------------
.................
4 0
3 years ago
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